The first common node of the two linked lists
The first public node of the two linked lists-Likou link
Intersecting Linked List-Likou Link
Topic content: 输入两个链表，找出它们的第一个公共节点。
Input and output example:
Code
Method 1: Use HashSet, make use of the non-repeatability of Hash, find it and return directly, and return to nothing after traversal without finding a hit
- The time and space complexity of this method are too high;
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
//思路一：采用HashSet，利用Hash的无重复性，找到直接返回，遍历完都没有找打，就返回空
Set<ListNode> set = new HashSet<>();
//1.首先将A放进去，在遍历B，(当然也可以反过来)
while (headA != null){
set.add(headA);
headA = headA.next;
}
//2.开始遍历B
while (headB != null){
if (set.contains(headB)){
return headB;//注意返回的是节点；
}
headB = headB.next;
}
return null;
}
Method 2: First calculate the length of the two linked lists, and then let the longer one go first, and when the remaining length is the same, then go together, traverse the past, find the common node, and return directly. After the traversal, it has not been found. , Can only return null
public static ListNode getIntersectionNode2(ListNode headA, ListNode headB) {
//思路二：先计算出俩个链表的长度，然后让长度比较长的先走，等剩余的长度一样长的时候，然后一起走，这样遍历过去，找到公共节点，直接返回,遍历完还没有找到，只能返回null
int lenA = length(headA);
int lenB = length(headB);
//1.首先让长的先走
while (lenA != lenB){
if (lenA > lenB){
headA = headA.next;
lenA--;
}else {
headB = headB.next;
lenB--;
}
}
//2.从循环中出来，长度一样，就一起走
while (headA != null){
//当然也可以是headB != null
if (headA.val == headB.val){
return headA;
}
headA = headA.next;
headB = headB.next;
}
return null;//遍历完还没有找到，只能返回空了
}
//计算长度的函数
public static int length(ListNode head){
int len = 0;
while (head != null){
len++;
head = head.next;
}
return len;
}
Method three:
public static ListNode getIntersectionNode3(ListNode headA, ListNode headB) {
//tempA和tempB我们可以认为是A,B两个指针
ListNode tempA = headA;
ListNode tempB = headB;
while (tempA != tempB) {
//如果指针tempA不为空，tempA就往后移一步。
//如果指针tempA为空，就让指针tempA指向headB（注意这里是headB不是tempB）
tempA = tempA == null ? headB : tempA.next;
//指针tempB同上
tempB = tempB == null ? headA : tempB.next;
}
//tempA要么是空，要么是两链表的交点
return tempA;
}